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# Evaluate $\int _ { C } \frac { x ^ { 2 } } { y ^ { 4 / 3 } } d s ,$ where $C$ is the curve $x = t ^ { 2 } , y = t ^ { 3 } ,$ for $1 \leq t \leq 2$

## $\frac{1}{27}\left(40^{3 / 2}-13^{3 / 2}\right)$

Integrals

Vectors

Vector Functions

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##### Lily A.

Johns Hopkins University

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

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### Video Transcript

Okay. What we want to go ahead and do is we want to evaluate, um, the integral over the curve of X squared over. Why, um, raised to the 4/3 d s. Um, where see is, um, the curve represented by X equal to t squared. And why equal to take cube and t is gonna go between two and a one and two inclusive. Okay, so the first thing we need to do ISS to defined our tea and so are of tea is gonna be t squared. I plus t cubed J. And so the derivative of our is the of tea, which is going to give me to t I plus three t squared J. Okay, Um and so, um, the magnitude of e of t is gonna be equal to the square root of two t squared plus a three t squared squared. So this is gonna give me a, um, square root of four t squared plus nine t to the fourth. Um, we're gonna factor out a T square to that's gonna give me a T times the square root of four plus nine t to the fourth t no t squared a factor Data T Square. So that's going to leave me with a T squared. Okay, so d s is equal to t times the square root of four plus nine t squared TT. Okay, um and so we've kind of got a good setup. Um, And so now let's sit up, are integral. So are integral. Um is going to go from 1 to 2, um, of x squared and x is t square to That's gonna be a t to the fourth over. Um, and why to the 4/3 which is going to give me a, um and why is thi cubes? So that is going to give me a T to the fourth times. T um, times four plus nine t squared. DT. Okay, so this could be the inner girl from 1 to 2 of tea times the square root of four plus 90 squared T t. Okay. And so once again, we're gonna do a use up. Ah, you substitution. I'ma let you be four plus 90 squared. So d'you is equal to 18 t d t. All I have is a d t d a t d t. So, um, 1/18 d'you is equal to T T t. So I'm gonna go ahead and do a substitution in there. So this becomes the integral, um, of or one over 18 times in nickel. And let's go ahead and change our upper and lower limits as well. So if t is one we get, um, you of 13. And if tea is too, um, we get 36 plus four, which gives me a you value of 40. Um And so this is going to be you to the 1/2. Do you? Which is gonna give me, um, won over 18 times. 2/3 you to the three have evaluated at 40 and then at 13. And so this is going to be won over 27 times 40 Raise to the three. Have minus 13 raised to the 3/2.

University of Central Arkansas

#### Topics

Integrals

Vectors

Vector Functions

##### Lily A.

Johns Hopkins University

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp